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(A) Let the four positive consecutive terms of the $G.P.$ be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ with common ratio $R = r^2$.Given product: $\frac{

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the four positive consecutive terms of the $G.P.$ be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ with common ratio $R = r^2$.Given product: $\frac{a}{r^3} 	imes \frac{a}{r} 	imes ar 	imes ar^3 = a^4 = 1296 \implies a = 6$.Given sum: $\frac{a}{r^3} + \frac{a}{r} + ar + ar^3 = 126$.Substituting $a=6$: $\frac{6}{r^3} + \frac{6}{r} + 6r + 6r^3 = 126 \implies (r^3 + \frac{1}{r^3}) + (r + \frac{1}{r}) = 21$.Let $x = r + \frac{1}{r}$. Then $r^3 + \frac{1}{r^3} = x^3 - 3x$.So,$(x^3 - 3x) + x = 21 \implies x^3 - 2x - 21 = 0$.By inspection,$x=3$ is a root: $27 - 6 - 21 = 0$.Thus,$r + \frac{1}{r} = 3 \implies r^2 - 3r + 1 = 0$.The common ratio is $R = r^2$. From $r^2 - 3r + 1 = 0$,we have $r^2 + 1 = 3r$.Dividing by $r$,$r + \frac{1}{r} = 3$. Squaring gives $r^2 + \frac{1}{r^2} + 2 = 9 \implies r^2 + \frac{1}{r^2} = 7$.The two possible common ratios are the roots of $t^2 - 7t + 1 = 0$ (since $R + \frac{1}{R} = 7$).The sum of the common ratios is $7$.

If the sum and product of four positive consecutive terms of a $G.P.$ are $126$ and $1296$ respectively,then the sum of common ratios of all such $G.P.s$ is $.........$.

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